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**Deon588****Member**- Registered: 2011-05-02
- Posts: 68

Hi all. I think i'm making a mistake somewhere as the matrix I get after multiplying doesn't seem to be invertible and the next question is to find A^-1. Is this the right way to do this?

Thanks in advance

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

I am getting:

which can be inverted.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Deon588****Member**- Registered: 2011-05-02
- Posts: 68

Hi Bobbym any idea where I made a mistake?

Thanks a lot

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

Your mistake is in line 3. Check how you multiplied those matrices.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

You multiplied the matrices incorrectly

**zetafunc.****Guest**

Never mind, bobbym was faster.

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi Deon588

Your inverses are correct, so then you need to do the multiplying.

(AB)C = A(BC) for matrices so you have a choice of which pair you multiply first. But you must preserve order.

eg if AB = D you then do DC not CD.

I'm going to do these both ways in case I choose the way you didn't (if you see what mean)

or

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

and you should find that inverse A = E3.E2.E1

This would be a good check that you have inv A right and that your multiplying is OK.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Deon588****Member**- Registered: 2011-05-02
- Posts: 68

Hi Bob thanks a lot for all the effort. I had no idea how to multiply more than 2 matrices so I tried doing all 3 at once...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi Deon588;

Always break a big problem down into smaller pieces that you know how to do. This top down design is very common in programming. You knew how to do 2 matrices.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi Deon588

I know the rule for multiplying two matrices. I dareay you could devise a rule for three but it would be hard to remember. So two at a time is simplest.

You shouldn't assume that matrices can be swopped around like numbers but associativity { (AB)C = A(BC) } does work.

Beware: commutativity doesn't. { AB ≠ BA }

Did you try my practice suggestion?

and you should find that inverse A = E3.E2.E1

This would be a good check that you have inv A right and that your multiplying is OK.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Deon588****Member**- Registered: 2011-05-02
- Posts: 68

Yes that was also the next question in the exam paper I did, worked out perfectly at the end

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

Bob

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